Transcript Slide 1

ECON6036
1st semester 05-06
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Format of final exam
Same as the mid term
Material not covered in final exam
fixed point theorem—both proof and application
purification of mixed strategy
rationalizability, dominance solvability
repeated game – overtaking criteria, limit of
means criteria, Maskin-Tirole theorem
• Cho-Kreps Intuitive criteria
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Material covered in final exam
strategic game
--Nash equilibrium
Bayesian game
--Bayesian equilibrium
extensive game with perfect information
--subgame perfect equilibrium
bargaining game
repeated game—subgame perfect equilibrium, trigger strategy,
minmax value
• extensive game with imperfect information
• --perfect Bayesian equilibrium
• --sequential equilibrium
Exercise 211.1 (Timing claims on
an investment)
• An amount of money accumulates; in period t (= 1, 2, ...,
T) its size is $2t.
• In each period t two people simultaneously decide
whether to claim the money. If only one person does so,
she gets all the money; if both people do so, they split
the money equally; either case, the game ends.
• If neither person does so, both people have the
opportunity to do so in the next period; if neither person
claims the money in period T, each person obtains $T.
• Each person cares only about the amount of money she
obtains.
• Formulate this situation as an extensive game with
perfect information and simultaneous moves, and find its
SPE.
Equilibrium: immediate claiming
• Claim: in the SPE, each player always claims money
whenever she is asked to move
• Proof: When t=T, it is the strictly dominant action for each
to claim (by claiming, she gets T rather than 0 if the other
also claims; she gets 2T rather than T if the other
doesn’t) => each always claims money at t=T
• Assume each always claims money at t=k+1,…,T
• Then at t=k, each claiming is also best response (by
claiming, she gets k rather than 0 if the other also claims;
she gets 2k rather than (k+1) if the other doesn’t).
• (There exists another SPE in which in the first period
neither claims money but in any subsequent period both
claim money.)
EXERCISE 227.3 (Sequential duel)
• In a sequential duel, two people alternately have the
opportunity to shoot each other; each has an infinite
supply of bullets.
• On each of her turns, a person may shoot or refrain from
doing so. Each of person i’s shots hits (and kills) its
intended target with probability p, (independently of
whether any other shots hit their targets).
• Each person cares only about her probability of survival
(not about the other person's survival).
• Model this situation as an extensive game with perfect
information and chance moves.
• Show that the strategy pairs in which neither person ever
shoots and in which each person always shoots are both
subgame perfect equilibria.
No shooting SPE
• Claim: each never shoots (whether or
not somebody has ever shot)
• Proof: According to the prescripts,
each’s survival probability is already
one, and cannot be further increased.
Hence, no beneficial unilateral
deviation.
Shooting SPE
• Claim: each always shoot
• Proof:
– We argue that deviating one is not beneficial.
Suppose now it is player 1’s turn to to move in period
t.
– Let Q<1 be 1’s payoff (survival probability) conditional
on both players exist in period t+1 and they act
according to the prescripts thereafter.
– If 1 shoots in period t and both act according to the
prescripts thereafter, 1’s payoff is p1+(1-p1)Q=p1(1Q)+Q. If he does not shoot in period t and both act
according to the prescripts, his payoff is Q. Clearly,
the deviation is NOT beneficial.
Example 473.1 (One-sided offers)
• Consider the variant of the bargaining game of
alternating offers in which only player 1 makes
proposals.
• In every period, player 1 makes a proposal,
which player 2 either accepts, ending the game,
or rejects, leading to the next period, in which
player 1 makes another proposal.
• Consider the strategy pair in which player 1
always proposes (x1,1-x1) and player 2 always
accepts a proposal (y1,y2) if and only if y2 ≥ 1-x1.
• Find the value(s) of x1 for which this strategy pair
is a subgame perfect equilibrium.
Equilibrium
• SPE: 1 always proposes (x1,1-x1); 2 always accepts a
proposal giving her at least 1-x1 and rejects any
inferior proposal.
• Claim: 1-x1=0.
• Proof: (use one stage deviation)
– Suppose not (so that 1-x1>0). Consider the history in which a
proposal (z1,1-z1) is proposed so that δ(1-x1)<1-z1<(1-x1).
– By accepting this offer, 2 gets 1-z1 now. By rejecting this offer,
2 will get δ(1-x1)< 1-z1.
– Hence, 2 should accept the proposal which is strictly inferior
than (x1,1-x1). But according to her prescript, she should not
accept such an inferior proposal. A contradiction.
EXERCISE 445.1 (Tit-for-tat as a
subgame perfect equilibrium)
• Consider the infinitely repeated
Prisoner's Dilemma in which the
payoffs of the component game
are those given in the Figure.
• Show that (tit-for-tat, tit-for-tat)
is a subgame perfect
equilibrium of this infinitely
repeated game with discount
factor δ if and only if y-x=1 and
δ= 1/x.
C
D
C
x,x
0,y
D
y,0
1,1
Note: 1 < x < y
Tit for Tat
• Tit-for-tat: do whatever the other did to you
in the previous period
• Four types of histories to check: those
ending with (C,C), (C,D), (D,C), (D,D).
• Need to show player 1 does not gain by one
deviation
(C,C)
Following equilibrium strategy, period outcomes will be (C,C), (C,C),...
and 1 will get
x
x (1      ...) 
1
If 1 deviates (to D), period outcomes alternate ./. (D,C) and (C,D) and
1 will get
y
2
4
y (1     ...) 
C
D
1 2
x
y
C
x,x 0,y
1 does not deviate if and only if

2
1 1
or (1   ) x  y.
D
y,0 1,1
2
C,D
By following equil strategy, the outcomes will alternate ./. (D,C) and (C,D)
and 1 will get
y
1 2
By deviating once (to C), the outcomes will become (C,C),(C,C),...
1 will get
x
x (1     2  ...) 
1
C
y
x
Hence 1 does not deviate if and only if

C x,x
1 2 1
or y  (1   ) x
y (1   2   4  ...) 
D
D
0,y
y,0 1,1
(D,C)
By following equil strategy, the outcomes will alternate ./. (C,D) and (D,C)
and 1 will get
y (   3  ...) 
y
1 2
By deviating once (to D), the outcomes will become (D,D),(D,D),...
1 will get
1
(1     2  ...) 
C
1
y
1
C x,x
Hence 1 does not deviate if and only if

2
1
1
or  y  (1   )
D
y,0
D
0,y
1,1
(D,D)
By following equil strategy, the outcomes will become (D,D),(D,D),...
1 will get
1
1
By deviating once (to C), the outcomes will alternate ./. (C,D) and (D,C)
and 1 will get
y
y (   3  ...) 
1 2
C
D
1
y
Hence 1 does not deviate if and only if

C x,x 0,y
1 1 2
or 1+   y
(1     2  ...) 
D
y,0
1,1
To summarize
• Conditions for no
deviations:
• CC:
y(1+δ)x
• CD:
y≥(1+δ)x
• DC:
δy≥1+δ
• DD:
δy1+δ
• Hence, y=(1+ δ)x and
δy=1+δ.
• Finally, δ=1/x and yx=1.
• Both using tit-for-tat is
SPE if and only if
these two conditions
hold.
• Very stringent
conditions indeed!
EXERCISE 331.1 (Selten's horse)
• Find the perfect Bayesian
equilibria of the game in
Figure 331.2 in which
each player's strategy is
pure.
• Hint: Find the pure
strategy Nash equilibria,
then determine which is
part of a weak sequential
equilibrium
c
d
C
1,1,1
4,4,0
D
3,3,2
3,3,2
L
c
d
C
1,1,1
0,0,1
D
0,0,0
0,0,0
R
• Two pure strat Nash equil: (D,c,L) and
(C,c,R)
• The 1st one is NOT part of a PBE.
Foreseeing 3 will choose L, 2 should
choose d to earn 4 rather than c to
earn 1. Not sequential rationality.
Hence not PBE.
• The 2nd one is part of a PBE.